The radius of Jupiter is 11 times that of Earth and the free fall
acceleration near its surface is 2.5 times that on earth. If we someday put
a spacecraft in low Jupiter orbit, its orbital speed will be :
a. greater than that for an Earth satellite (Assume low orbit)
b. same as that for an Earth satellite (Assume low orbit)
c. Less than that for an Earth satellite (Assume low orbit)

There is a 446N force in the horizontal massless cable CD. The vertical component of the pivot A's force on the aluminum pole, which is 328.3N. The force that the pivot A applies to the aluminum pole has a horizontal component of 446N.

We need to be aware of the force in order to discover the solution.

where, H=3.8m, l=7.6m, m=10kg, M=23.5 kg and alpha= 37 degree,

                      [tex]F_V-mg-Mg=0\\F_v=328.2N\\[/tex]

                  [tex]F_H=T=446N[/tex]

In light of this, we may say that the tension in the horizontal massless cable CD and the horizontal component of the force applied by the pivot A to the aluminum pole are identical and each exert 466N of force, whereas the vertical component of that force is 328.3N.

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(a) The maximum height reached by Jane is 1.8 m.

(b) The length of the vine will affect the time of her motion, which will impact on speed and maximum height attained.

apply the principle of conservation of energy;

P.E = K.E

mgh = ¹/₂mv²

h = v²/2g

where;

h = (6²)/(2 x 9.8)

h = 1.84 m

Assuming Jane to be in simple harmonic motion, the time of motion is calculated as;

T =  2π√(L/g)

where;

Thus, the length of the vine will affect the time of her motion, which will impact on speed and maximum height attained.

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Answer:

D. Impulse

Explanation: Hope this helps

Answer:

See below

Explanation:

I will assume the force is in a DOWNWARD direction ( I believe it makes no answer difference)

Horizontal component is then 100 cos 36° =80.9 N

F = ma

80.9 = 25 kg  *a

a = 3.24 m/s^2

Answer:

See image

Explanation:

Plato

Answer:

No

Explanation:

Plastic bags are not magnetic materials, only magnetic materials (such as iron) can be attracted by the magnet.

Hope this helps.

Answer:

Approximately [tex]25\; {\rm N}[/tex] assuming that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].

Explanation:

This mass is in a circular motion of radius [tex]r[/tex]. Hence, when the velocity of the mass is [tex]v[/tex], the acceleration of this mass should be [tex](v^{2} / r)[/tex]. The net force on this mass should be [tex](\text{net force}) = (m\, v^{2}) / r[/tex] towards the center of the circle.

When this [tex]m = 0.80\; {\rm kg}[/tex] mass is at the top of this circle, both gravitational pull and the force of the string (tension) point downwards. Hence, the net force on this mass would be:

[tex](\text{net force}) = (\text{weight}) + (\text{tension})[/tex].

Thus:

[tex]\begin{aligned} (\text{tension}) &= (\text{net force}) -(\text{weight})\\ &= \frac{m\, v^{2}}{r} - m\, g \\ &= m\, \left(\frac{v^{2}}{r} - g\right) \\ &= 0.80\; {\rm kg}\times \left(\frac{(9.0\; {\rm m\cdot s^{-1}})^{2}}{2.0\; {\rm m}} - 9.81\; {\rm m\cdot s^{-2}}\right) \\ &\approx 25\; {\rm kg \cdot m\cdot s^{-2}} \\ &= 25\; {\rm N}\end{aligned}[/tex].

The magnitude of the cannons velocity after the ball is fired will be 2.4m/s.

To find the answer, we have to know more about the law of conservation of linear momentum.

                      [tex]M_B=8kg\\M_C=1*10^3 kg\\P_f=2.4*10^3 kgm/s.\\[/tex]

                   [tex]P_f=M_CV_C\\V_C=\frac{P_f}{M_C}=\frac{2.4*10^3}{1*10^3} =2.4m/s \\west[/tex]

Thus, we can conclude that, the magnitude of the cannons velocity after the ball is fired is 2.4m/s.

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After the ball is shot, the cannon will move at a speed of 2.4 m/s.

We must learn more about the rule of conservation of linear momentum in order to locate the solution.

How can I determine the cannon's post-ball velocity magnitude?

                          [tex]m=8kg\\M=1000kg\\P_f=2.4*10^3kgm/s[/tex]

                               [tex]P_f=MV\\V=\frac{P_f}{M} =2.4m/s[/tex]

Thus, we can infer that the cannon's maximum speed once the ball is fired is 2.4 m/s.

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The thermal energy that is generated due to friction is 344J.

Now we know that the total mechanical energy in the system is constant. The loss in energy is given by the loss in energy.

Thus, the kinetic energy is given as;

KE = 0.5 * mv^2 =0.5 * 15.0-kg * (1.10 m/s)^2 = 9.1 J

PE = mgh = 15.0-kg * 9.8 m/s^2 *  2.40 m = 352.8 J

The thermal energy is; 352.8 J - 9.1 J = 344J

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Answer:

Explanation:

The angular speed of the rotor is 200 rad/s.

The torque needed to be transmitted by the engine is 180 Nm.

The power of the rotor required to transmit energy to apply a torque τ to rotate a motor with angular speed ω,

P=τω

=180×200W

=36kW

The highest point of the wheel is the position of the wheel when its potential energy is greatest.

The position of the wheel when its potential energy is greatest when it is at the highest point because potential energy depends on the height of an object. If the object is at more height then it has more potential energy and vice versa.

So we can conclude that the highest point of the wheel is the position of the wheel when its potential energy is greatest.

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The tension in the upper rope is 50.53 N.

The tension in the upper rope is calculated as follows;

T(up) = T(dn) + mg

where;

T(up) = 12.8 N + (3.85 x 9.8) N

T(up) = 50.53 N

Thus, the tension in the upper rope is 50.53 N.

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C. The opportunity cost of a student who is staying up all night to study for an exam that he has to take in the early morning is sleep or rest.

The law of increasing opportunity cost is an economic principle that describes how opportunity costs increase as resources are applied.

As the student gives up his sleep or night rest in the place of his exam preparation, we say that the opportunity cost is the sleep or rest.

Thus, the opportunity cost of a student who is staying up all night to study for an exam that he has to take in the early morning is sleep or rest.

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(1) The speed of light through vacuum is 3 x 10⁸ m/s.

(2) The speed of light in air at 30⁰C is 350 m/s.

(3) The speed of light in air at 0⁰C is 330 m/s.

(4) The transparent substance is glass.

(5) The color of light observed is red.

(6) The energy of a photon of red light is 2.85 x 10⁻¹⁹ J.

The speed of light through vacuum is 3 x 10⁸ m/s.

The speed of light in air at 30⁰C is 350 m/s.

The speed of light in air at 0⁰C is 330 m/s

n = speed of light in vacuum/speed of light in the substance

n = ( 3 x 10⁸ m/s) / ( 2 x 10⁸ m/s)

n = 1.5

The refractive index of glass is 1.5, thus, the transparent substance is glass.

f = v/λ

f = (3 x 10⁸) / (6.8 x 10⁻⁷)

f = 4.41 x 10¹⁴ Hz

The color of light observed is red based on the frequency and wavelength.

E = hf

E = (6.626 x 10⁻³⁴)(4.3 x 10¹⁴)

E = 2.85 x 10⁻¹⁹ J

Thus, the speed of light through vacuum is 3 x 10⁸ m/s.

The speed of light in air at 30⁰C is 350 m/s.

The speed of light in air at 0⁰C is 330 m/s.

The transparent substance is glass.

The color of light observed is red.

The energy of a photon of red light is 2.85 x 10⁻¹⁹ J.

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The maximum height h of the curved path is 7.38 m.

Apply the following kinematic equation;

v² = u² - 2gh

where;

(u² - v²)/2g = h

(39² - 37.1²)/(2 x 9.8) = h

7.38 m = h

Thus, the maximum height h of the curved path is 7.38 m.

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The angle of the red light is mathematically given as

[tex]\theta = 7.24 \textdegree[/tex]

Generally, the equation for the grating element is  mathematically given as

d= 1 / N

Therefore

d= 1/1965

d= 5.089 * 10^{-6} m

Generally, the equation for the  differential formula is  mathematically given as

[tex]d sin \theta = m\lambda[/tex]

Therefore

[tex]sin \theta = \lambda / d[/tex]

[tex]sin \theta= (640 * 10 ^ {-9} m)/(5.089 * 10 ^ {-6} m)[/tex]

[tex]\theta = 7.24 \textdegree[/tex]

In conclusion, The angle of the red light

[tex]\theta = 7.24 \textdegree[/tex]

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Hi there!

We can use Biot-Savart's Law for a moving particle:
[tex]B= \frac{\mu_0 }{4\pi}\frac{q\vec{v}\times \vec{r}}{r^2 }[/tex]

B = Magnetic field strength (T)
v = velocity of electron (0.130c = 3.9 × 10⁷ m/s)

q = charge of particle (1.6 × 10⁻¹⁹ C)

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

r = distance from particle (2.10 μm)

There is a cross product between the velocity vector and the radius vector (not a quantity, but specifies a direction). We can write this as:

[tex]B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }[/tex]

Where 'θ' is the angle between the velocity and radius vectors.

a)
To find the angle between the velocity and radius vector, we find the complementary angle:

θ = 90° - 60° = 30°

Plugging 'θ' into the equation along with our other values:

[tex]B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }\\\\B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(30)}{(2.1*10^{-5})^2 }[/tex]

[tex]B = \boxed{7.07 *10^{-10} T}[/tex]

b)
Repeat the same process. The angle between the velocity and radius vector is 150°, and its sine value is the same as that of sin(30°). So, the particle's produced field will be the same as that of part A.

c)

In this instance, the radius vector and the velocity vector are perpendicular so

'θ' = 90°.

[tex]B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(90)}{(2.1*10^{-5})^2 } = \boxed{1.415 * 10^{-9}T}[/tex]

d)
This point is ALONG the velocity vector, so there is no magnetic field produced at this point.

Aka, the radius and velocity vectors are parallel, and since sin(0) = 0, there is no magnetic field at this point.

[tex]\boxed{B = 0 T}[/tex]

A. The magnitude of the spring force (in N) acting upon the object is 15.9 N

B. The magnitude of the object's acceleration (in m/s²) is 30.58 m/s²

C. The direction of the acceleration vector points toward the equilibrium position (i.e., to the left in the figure).

F = Ke

F = 106 × 0.15

F = 15.9 N

F = ma

Divide both sides by m

a = F / m

a = 15.9 / 0.52

a = 30.58 m/s²

Considering the diagram, we can see that the spring was pulled away from the equilibrium point.

Thus, when the spring is released, it will move toward the equilibrium point. This is also true about the acceleration.

Therefore, we can conclude that the direction of the acceleration vector is towards the equilibrium point.

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The spring constant, k, in newtons per meter is 1,955.9 N/m.

h = v²sin²θ/2g

v = √[(2gh)/sin²θ]

v = √[(2 x 9.8 x 4.4)/ (sin 39)²]

v = 14.76 m/s

E = K.E + P.E

E = ¹/₂m(v cosθ)²  +  mgh

E =  ¹/₂(1.1)(14.76 cos39)²  +  (1.1 x 9.8 x 4.4)

E = 119.8 J

E = ¹/₂kx²

k = 2E/x²

k = (2 x 119.8)/(0.35²)

k = 1,955.9 N/m

Thus, the spring constant, k, in newtons per meter is 1,955.9 N/m.

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(a) The initial speed of the rock as it left the astronaut's hand is 19.35 m/s.

(b) The speed of the satellite is 50.24 m/s.

g = GM/R²

where;

g = (6.67 x 10⁻¹¹ x 7.88 x 10¹⁸)/(63200)²

g = 0.13 m/s²

v² = u² - 2gh

where;

at maximum height, v = 0

u² = 2gh

u = √2gh

u = √(2 x 0.13 x 1,440)

u = 19.35 m/s

v = √GM/r

r = radius of planet Glob + radius of the satellite

r = 63200 m + 145,000 m = 208,200 m

v = √[(6.67 x 10⁻¹¹ x 7.88 x 10¹⁸)/(208,200)]

v = 50.24 m/s

Thus, the initial speed of the rock as it left the astronaut's hand is 19.35 m/s.

The speed of the satellite is 50.24 m/s.

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Answer:

Explanation:

the value is -8 cm

The approximate uncertainty in the area of a circle is 5.4%.

A = πr²

where;

A = π(4.3 x 10⁴)²

A = 5.81 x 10⁹ cm²

Let the error in measurement = 1 x 10⁴ cm

A =  π(1 x 10⁴)² = 3.14 x 10⁸ cm²

% = (error/actual area) x 100%

% = ( 3.14 x 10⁸) / (5.81 x 10⁹) x 100%

% = 5.4 %

Thus, the approximate uncertainty in the area of a circle is 5.4%.

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Explanation:

gravitational potential energy = mgh (must be in S.I. unit)

m= 0.4 kg ; g= 10m/s (gravitational acceleration occurs); h=9.2 m

hence mgh=0.4×10×9.2= 36.8J

unit for energy is joules and since the variables are in S.I. unit, we can use Joules as the final unit for measurement

Answer:

36.06 J if you define g = 9.8 m/s² or 36.8 J if you define g = 10 m/s²

Explanation:

Potential Energy = Mass x Gravitational Acceleration x Height

It can be expressed as [tex]\displaystyle{PE = mgh}[/tex] where PE is potential energy, m is mass, g is gravitational acceleration and h is height.

In this case, we know mass is 0.40 kg and height of 9.2 m as well as gravitational acceleration is defined to be 9.8 m/s² (You can also define g = 10 m/s²)

Therefore, substitute given information in formula:

[tex]\displaystyle{PE=0.40 \ \times \ 9.8 \ \times 9.2 }\\\\\displaystyle{PE=36.06 \ J}[/tex]

With g = 10 m/s², you'll get:

[tex]\displaystyle{PE = 0.40 \ \times 10 \ \times 9.2}\\\\\displaystyle{PE = 36.8 \ J}[/tex]

Note that J is for joule unit.

Therefore, the answer is 36.06 J if you define g = 9.8 m/s² or 36.8 J if you define g = 10 m/s² - both work.

Answer:

Create space

Watch your breath

Watch your thoughts

Be gentle with yourself

Affirm what you want

Explanation:

Answer:

Hola como estás ehord as ve hi5 ido

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fje di oo0008t aquí 1gvu txdc

Since upward forces must be equal to the downward forces, the tension in the cable is 225.3 N

The two conditions are;

The given parameters are;

Tsinθ = mg

Tsin 66 = 21 x 9.8

T = 205.8 / 0.914

T = 225.3 N

Therefore, the tension in the cable is 225.3 N

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The solution for the three questions  is mathematically given as

Parts A and C are both zeros.

Part B B = 0.001855Tesla

Parts A and C are both zeros.

For component A, the magnetic field is zero since 12.6 cm is still inside the toroidal solenoid. Part C has no magnetic field since it is 20.7 cm outside of the toroidal solenoid.

Generally, the equation for  magnetic field is  mathematically given as

B = (mu_0*N*I)/(2*pi*r)

Therefore

B = ((4*pi*10^{-7})*180*8.40)/(2*pi*0.163)

B = 0.001855Tesla

In conclusion, the magnitude of the magnetic field at 16.3 cm from the center of the torus is

B = 0.001855Tesla

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Answer:

Part A & C: 0 T

Part B: 1.855*10^-3 T

Explanation:

The formula that models the magnetic field of a toroid is B=μ0*N*I/2π*r .

Note: Toroids keep their magnetic field rotating within the coils that carry current.

Part A & C: Thus the B field magnitude 12.6 cm & 20.7 cm away from the center is 0 T.

Part B: [tex]B=\frac{(4\pi *10x^{-7} )(180)(8.4 A)}{(2\pi )(16.3*10x^{-2} m)}[/tex]

B=1.855*10^-3 T

(a) The proton’s potential energy change is 3.6 x 10⁻¹⁸ J.

(b) The potential difference between the negative plate and a point midway between the plates is 11.25 V.

(c) The speed of the proton just before it hits the negative plate is 6.57 x 10⁴ m/s.

U = qΔV

where;

U = q(Ed)

U = (1.6 x 10⁻¹⁹)(1500 x 1.5 x 10⁻²)

U = 3.6 x 10⁻¹⁸ J

ΔV = E(0.5d)

ΔV = 0.5Ed

ΔV = 0.5 (1500)(1.5 x 10⁻²)

ΔV = 11.25 V

U = ¹/₂mv²

U = mv²

v² = 2U/m

where;

v² = (2 x 3.6 x 10⁻¹⁸) / ( 1.67 x 10⁻²⁷)

v² = 4.311 x 10⁹

v = √(4.311 x 10⁹)

v = 6.57 x 10⁴ m/s

Thus, the proton’s potential energy change is 3.6 x 10⁻¹⁸ J.

The potential difference between the negative plate and a point midway between the plates is 11.25 V.

The speed of the proton just before it hits the negative plate is 6.57 x 10⁴ m/s.

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(A) The spring constant of the spring is 4.94 N.

(B) The speed of the cart after pushing it is 0.47 m/s.

(C) The force applied to the cart is 0.75 N.

ω = √k/m

where;

0.5 rev/s = 0.5(2π) rad/s = π rad/s = 3.142 rad/s

ω² = k/m

k = mω²

k = 0.5 x (3.142)²

k = 4.94 N/m

E = ¹/₂kA²

where;

A is amplitude

E = ¹/₂(4.94)(0.15)²

E = 0.056 J

E = ¹/₂mv²

2E = mv²

v² = 2E/m

v² = (2 x 0.056)/(0.5)

v² = 0.224

v = √0.224

v = 0.47 m/s

E = ¹/₂FA

2E = FA

F = 2E/A

F = (2 x 0.056)/(0.15)

F = 0.75 N

Thus, the spring constant of the spring is 4.94 N. The speed of the cart after pushing it is 0.47 m/s. The force applied to the cart is 0.75 N.

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